Given that $f(x) = x^k$ where $k > 0$, what is the range of $f(x)$ on the interval $[1, \infty)$?
Solution: Because $k > 0$, $f(x)$ is increasing on the interval $[1, \infty)$. We see that $f(1) = 1^k = 1$, and as $x$ increases, $f(x) = x^k$ increases without bound.  Hence, on the interval $[1,\infty)$, $f(x)$ takes on all values greater than or equal to 1, which means the range of $f(x)$ is $\boxed{[1,\infty)}$.